%I
%S 1,0,2,1,6,7,22,36,89,168,377,756,1630,3353,7110,14783,31130,65016,
%T 136513,285648,599041,1254456,2629418,5508097,11542854,24183271,
%U 50674318,106173180,222470009,466131960,976694489,2046447180,4287928678,8984443769,18825088134
%N Constant term of the reduction of nth Fibonacci polynomial by x^2 > x+1. (See Comments.)
%C Polynomial reduction: an introduction
%C ...
%C We begin with an example. Suppose that p(x) is a polynomial, so that p(x)=(x^2)t(x)+r(x) for some polynomials t(x) and r(x), where r(x) has degree 0 or 1. Replace x^2 by x+1 to get (x+1)t(x)+r(x), which is (x^2)u(x)+v(x) for some u(x) and v(x), where v(x) has degree 0 or 1. Continuing in this manner results in a fixed polynomial w(x) of degree 0 or 1. If p(x)=x^n, then w(x)=x*F(n)+F(n1), where F=A000045, the sequence of Fibonacci numbers.
%C In order to generalize, write d(g) for the degree of an arbitrary polynomial g(x), and suppose that p, q, s are polynomials satisfying d(s)<d(q). By the division algorithm, there exists a unique pair t and r of polynomials such that p=q*t+r and d(r)<d(q). Replace q by s to get s*t+r, which is q*u+v for some u and v, where d(v)<d(q). Continue applying q>s in this manner until reaching w such that d(w)<d(q). We call w the reduction of p by q>s.
%C The coefficients of (reduction of p by q>s) comprise a vector of length d(q)1, so that a sequence p(n,x) of polynomials begets a sequence of vectors, such as (F(n), F(n1)) in the above example. We are interested in the component sequences (e.g., F(n1) and F(n)) for various choices of p(n,x).
%C Following are examples of reduction by x^2>x+1:
%C nth Fibonacci p(x) > A192232+x*A112576
%C nth cyclotomic p(x) > A192233+x*A051258
%C nth 1stkind Chebyshev p(x) > A192234+x*A071101
%C nth 2ndkind Chebyshev p(x) > A192235+x*A192236
%C x(x+1)(x+2)...(x+n1) > A192238+x*A192239
%C (x+1)^n > A001519+x*A001906
%C (x^2+x+1)^n > A154626+x*A087635
%C (x+2)^n > A020876+x*A030191
%C (x+3)^n > A192240+x*A099453
%C ...
%C Suppose that b=(b(0), b(1),...) is a sequence, and let p(n,x)=b(0)+b(1)x+b(2)x^2+...+b(n)x^n. We define (reduction of sequence b by q>s) to be the vector given by (reduction of p(n,x) by q>s), with components in the order of powers, from 0 up to d(q)1. For k=0,1,...,d(q)1, we then have the "ksequence of (reduction of sequence b by q>s)". Continuing the example, if b is the sequence given by b(k)=1 if k=n and b(k)=0 otherwise, then the 0sequence of (reduction of b by x^2>x+1) is (F(n1)), and the 1sequence is (F(n)).
%C ...
%C For selected sequences b, here are the 0sequences and 1sequences of (reduction of b by x^2>x+1):
%C b=A000045, Fibonacci sequence (1,1,2,3,5,8,...) yields
%C 0sequence A166536 and 1sequence A064831.
%C b=(1,A000045)=(1,1,1,2,3,5,8,...) yields
%C 0sequence A166516 and 1sequence A001654.
%C b=A000027, natural number sequence (1,2,3,4,...) yields
%C 0sequence A190062 and 1sequence A122491.
%C b=A000032, Lucas sequence (1,3,4,7,11,...) yields
%C 0sequence A192243 and 1sequence A192068.
%C b=(A000217, triangular sequence (1,3,6,10,...) yields
%C 0sequence A192244 and 1sequence A192245.
%C b=(A000290, squares sequence (1,4,9,16,...) yields
%C 0sequence A192254 and 1sequence A192255.
%C More examples: A192245A192257.
%C ...
%C More comments:
%C (1) If s(n,x)=(reduction of x^n by q>s) and
%C p(x)=p(0)x^n+p(1)x^(n1)+...+p(n)x^0, then
%C (reduction of p by q>s)=p(0)s(n,x)+p(1)s(n1,x)
%C +...+p(n1)s(1,x)+p(n)s(0,x). See A192744.
%C (2) For any polynomial p(x), let P(x)=(reduction of p(x)
%C by q>s). Then P(r)=p(r) for each zero r of
%C q(x)s(x). In particular, if q(x)=x^2 and s(x)=x+1,
%C then P(r)=p(r) if r=(1+sqrt(5))/2 (golden ratio) or
%C r=(1sqrt(5))/2.
%H Vincenzo Librandi, <a href="/A192232/b192232.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,1,1).
%F Empirical G.f.: x*(x^2+x1)/(x^4+x^33*x^2x+1).  _Colin Barker_, Sep 11 2012
%F The above formula is correct.  _Charles R Greathouse IV_, Jan 08 2013
%F a(n) = A265752(A206296(n)).  _Antti Karttunen_, Dec 15 2015
%F a(n) = A112576(n) A112576(n1) A112576(n2).  _R. J. Mathar_, Dec 16 2015
%e The first four Fibonacci polynomials and their reductions by x^2>x+1 are shown here:
%e F1(x)=1 > 1 + 0x
%e F2(x)=x > 0 + 1x
%e F3(x)=x^2+1 > 2+1x
%e F4(x)=x^3+2x > 1+4x
%e F5(x)=x^4+3x^2+1 > (x+1)^2+3(x+1)+1 > 6+6x.
%e From these, read A192232=(1,0,1,1,6,...) and A112576=(0,1,1,4,6,...).
%t q[x_] := x + 1;
%t reductionRules = {x^y_?EvenQ > q[x]^(y/2), x^y_?OddQ > x q[x]^((y  1)/2)};
%t t = Table[FixedPoint[Expand[#1 /. reductionRules] &, Fibonacci[n, x]], {n, 1, 40}];
%t Table[Coefficient[Part[t, n], x, 0], {n, 1, 40}]
%t (* A192232 *)
%t Table[Coefficient[Part[t, n], x, 1], {n, 1, 40}]
%t (* A112576 *)
%t (* _Peter J. C. Moses_, Jun 25 2011 *)
%t LinearRecurrence[{1, 3, 1, 1}, {1, 0, 2, 1}, 60] (* _Vladimir Joseph Stephan Orlovsky_, Feb 08 2012 *)
%o (PARI) Vec((1xx^2)/(1x3*x^2+x^3+x^4)+O(x^99)) \\ _Charles R Greathouse IV_, Jan 08 2013
%Y Cf. A168561, A192233  A192240, A192744.
%Y Cf. A206296, A265398, A265399, A265752, A265753.
%K nonn,easy
%O 1,3
%A _Clark Kimberling_, Jun 26 2011
%E Example corrected by _Clark Kimberling_, Dec 18 2017
